A possible example
I suspect there is no scheme which is "two $\mathbb A^1$'s glued together along their generic points" (or "$\mathbb A^1$ with every closed point doubled"). In other words, I think that the coequalizer of the two inclusions $Spec(k(t))\rightrightarrows \mathbb A^1\sqcup \mathbb A^1$ does not exist in the category of schemes. Intuitively, this coequalizer should be "too non-separated" to be a scheme.
I don't have a proof, but I thought other people might have ideas if I posted this here.
One thing I can prove is that if a coequalizer $X$ does exist, then no two closed points of $\mathbb A^1\sqcup \mathbb A^1$ map to the same point in $X$. To show this, it is enough to find functions from $\mathbb A^1\sqcup \mathbb A^1$ to other schemes which agree on the generic points but disagree on any other given pair of points. The obvious map $\mathbb A^1\sqcup \mathbb A^1\to \mathbb A^1$ separates most pairs of closed points. To see that a point on one $\mathbb A^1$ is not identified with "the same point on the other $\mathbb A^1$", consider the map from $\mathbb A^1\sqcup \mathbb A^1$ to $\mathbb A^1$ with the given point doubled.
Here are some questions that might be helpful to answer:
If the coequalizer above does exist, must the map from $\mathbb A^1\sqcup \mathbb A^1$ be surjective?
(see the related question Can a coequalizer of schemes fail to be surjective?)
Is the coequalizer of $Spec(k(t))\rightrightarrows \mathbb A^1\sqcup \mathbb A^1$ in the category of separated schemes equal to $\mathbb A^1$? (probably)